Possibly under certain conditions, but not generally.
Consider a nonmeasurable set A, and define
f(x) = 1 if x in A
0 otherwise.
Then {1} is certainly measurable but the inverse image
{x | f(x) = 1} = A is not measurable.
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Yes.
Inverse operations. Additive inverse is not one operation but they are elements of a set.
Given a set and a binary operation defined on the set, the inverse of any element is that element which, when combined with the first, gives the identity element for the binary operation. If the set is integers and the binary operation is addition, then the identity is 0, and the inverse of an integer k is -k. If the set is rational numbers and the binary operation is multiplication, then the identity element is 1 and the inverse of any member of the set, x (other than 0) is 1/x.
One way is to check whether the pre-image of the product is sigma-algebra. Please list an example to clarify your question.
It gives closure to the set of real numbers with regard to the binary operation of addition. This makes the set a ring. The additive inverse is used, sometimes implicitly, in subtraction.